Staff: Mentor

If you plot 1/acceleration versus speed, your function value is proportional to the time spent at a specific velocity (assuming your acceleration is always positive).
If you plot speed/acceleration versus speed, the area below the function is proportional to the distance travelled.

The area under the v/t graph should give total distance, whatever the instantaneous velocity and accelerations are. Does that hold for an (a/v)/v graph too? Is there not a constant needed after the integration?

Staff: Mentor

A simple example: v(t)=at+c with constant a and c.
Distance after time T is ##\frac{1}{2}aT^2 + cT##.

v/a = t + c/a
t=(v-c)/a
Therefore, v/a = (v-c)/a + c/a = v/a (trivial in this example, as a is constant)
Speed increases from c to c+aT. If we integrate v/a from c to c+aT, we get ##\int \frac{v}{a}dv = \frac{1}{2a}((c+aT)^2-c^2) = cT+\frac{1}{2}aT^2## - the same as above.

Wine and gin make this less accessible but the sums seem right in this case. It that also OK for non constant acceleration? I guess it could be - piecewise. Good one. I love it when someone else does the algebra and I can follow it.

Staff: Mentor

If you plot 1/acceleration versus speed, your function value is proportional to the time spent at a specific velocity (assuming your acceleration is always positive).
If you plot speed/acceleration versus speed, the area below the function is proportional to the distance travelled.

Not just proportional to the distance traveled. It is equal to the distance traveled. Another way to do this is to plot 1/(2a) as a function of v2. This will also be equal to the distance traveled.